Non-zero eigenvalues of $AA^T$ and $A^TA$ As a part of an exercise I have to prove the following:
Let $A$ be an $(n \times m)$ matrix. Let $A^T$ be the transposed matrix of $A$. Then $AA^T$ is an $(n \times n)$ matrix and $A^TA$ is an $(m \times m)$ matrix. $AA^T$ then has a total of $n$ eigenvalues and $A^TA$ has a total of $m$ eigenvalues.
What I need to prove is the following:
$AA^T$ has an eigenvalue $\mu \not = 0$ $\Longleftrightarrow$ $A^TA$ has an eigenvalue $\mu \not = 0$
In other words, they have the same non-zero eigenvalues, and if one has more eigenvalues than the other, then these are all equal to $0$.
How can I prove this?
Thanks and regards.
 A: Let $\lambda$ be an eigenvalue of $A^TA$, i.e. $$A^T A x = \lambda x$$
for some $x \neq 0$. We can multiply $A$ from the left and get
$$A A^T (Ax) = \lambda (Ax).$$
What can you conclude from this?
A: One proof that comes to mind is to use Sylvester's determinant theorem.
In particular:
$$
\mu \neq 0 \text{ is an eigenvalue of }A^TA  \implies\\
\det(A^TA - \mu I) = 0 \implies\\
\det(I + (-1/\mu)A^TA) = 0 \implies\\
\det(I + A(-1/\mu)A^T) = 0 \implies\\
\det(AA^T - \mu I) = 0 \implies\\
\mu \neq 0 \text{ is an eigenvalue of }AA^T
$$
A: SVD is definitely an overkill, but maybe it would be helpful to you (as it is for me) to draw the matrices that the decomposition gives us:

Recall that:


*

*The columns of $V$ (right-singular vectors) are eigenvectors of
$A^TA$.

*The columns of $U$ (left-singular vectors) are eigenvectors
of $AA^T$.

*$V^T=V^{-1}$ and $U^T=U^{-1}$.


With some simple operations you can get:

And also:

Write down both of these equations, but for the $i^\text{th}$ column.
From there, the solution is not far.
A: in fact, nonzero eigenvalues $AB$ and $BA$ are the same for any rectangular matrices
$A$ and $B$. this follows from the fact that $trace((AB)^k) = trace((BA)^k)$ and the coefficients of the characteristic polynomials of a square matrix $A$ are a function of $trace(A^k).$ 
